How to get a unit vector from another unit vector and angle between them? How to get a unit vector from another unit vector and angle between them? Is it possible?
I need something like  this:

 A: You can find the resulting vector by using a rotation matrix.
Alternatively, the unit circle $x^2 + y^2 = 1$ (the set of all vectors of length one) can be parameterised by $x = \cos\theta$, $y = \sin\theta$ where $\theta$ is the anti-clockwise angle made with the positive $x$-axis. As you know the angle the desired vector makes with the positive $x$-axis, you have the value $\theta$.
It is worth noting that neither method relies on the vectors being unit length.
A: Rotate the original vector by the angle $\theta$ specified, and divide it by its norm. If the original vector is $\vec{u}$ then you want a vector $\vec{v}$ given by
$$\vec{v}=R(\theta)\frac{ \vec{u}}{|\vec{u}|} $$
A: Let $\,(a,b)\,$ be the wanted vector, then;
$$\frac{\sqrt 3}{2}=\cos 30^\circ=\frac{(a,b)\cdot(0,1)}{||(a,b)||\,||(0,1)||}=\frac{b}{\sqrt{a^2+b^2}}\implies$$
$$4b^2=3a^2+3b^2\iff b^2=3a^2$$
You have, of course, several options to choose from...
A: You can just multiply it (if you use column vectors, otherwise transpose it before and after) by a Rotation matrix $$\begin {pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end {pmatrix}$$  But why do you show a horizontal vector as $(0,1)?$  It should be $(1,0)$  This would give the new vector as $(\cos 30^\circ, \sin 30^\circ)$
